Question 3 by Looi Wei Chern

Question for discussion

Based on the above conversation discuss, with examples and justification whether the following statement is        justified.

'A square is a rhombus but a rhombus is not a square'.

A rhombus is a quadrilateral whose four sides are all congruent    A square is a quadrilateral whose four sides are all congruent and whose angles are all right angles  In other words,    A square is a rhombus that is also a rectangle  These are both rhombuses:    +--------+   |        |        +--------+   |        |       /        /   |        |      /        /   |        |     /        /   +--------+    +--------+  But only this is a square:    +--------+   |        |   |        |   |        |   |        |   +--------+ 

Question 2 by Looi Wei Chern (Images)

1st Image is for Evidence for A

2nd Image is for Evidence for B

3rd Image is for Evidence for C

Question 2 by Looi Wei Chern

Question 2:

Which of the given statements is correct? Justify your answer/s with examples.

A ) A square and a parallelogram are quadrilaterals.

B ) Opposite sides of a square and a parallelogram are parallel.

C ) A trapezoid has one pair of parallel sides.

D ) All the above

D is Correct.

Evidence for A) A square and a parallelogram are both quadrilaterals as the definition of a quadrilateral is a polygon with four sides and four vertices or corners. Thus, the following are all quadrilaterals, which includes squares and parallelograms:

                                       

 

Evidence for B) Parallel lines are two lines lying in the same plane but never meeting no matter how far extended as it can be clearer seen below in the pictures of the square and the parallelogram.

                                       

Evidence for C) Like I said above, the definition of parallel lines is two lines lying in the same plane but never meeting no matter how far extended thus from the diagram below, only DC and AB is parallel to one another but AD and BC are not parallel to one another thus the definition of a trapezoid simply means a four-sided figure with one pair of parallel sides.

103's CLASSIFICATION OF QUADRILATERALS

Summary of what we have learnt - T3 W03 (12 July - 16 July)

• When we are factorising expressions by grouping we need to take care of the signs ... for instance ... 1 + 2t - k - 2kt = (1+2t) - k(1 + 2t) ... the LEFT hand term (1 + 2t) is the same as 1(1 + 2t) ... hence, when we factorise the expression, there is a COMMON factor of (1+2t) ... leaving us with ... (1+2t)(1 - k) ... some of you have forgotten the (1) in the term (1 - k) !!!
• When we work with algebraic fractions, make sure that you FACTORISE both the NUMERATOR and DENOMINATOR !!!

Term 3 Week 2 - 5th to 9th July 2010

Most of the class can understand what was being taught today, but some still have doubts about the equation "undefined" .

QUESTION -

Why 2/0 is undefined when 2÷0=0?

Well, firstly, note that 2/0 is the SAME as 2÷0 & hence, it would still be UNDEFINED!

Anyway, if 2 ÷ 0 is possible, we will be able to see that

letting 2 ÷ 0 = x

0 ( 2 ÷ 0) = 0( x ) .................................................................... convince yourself that 3( x ÷ 3) = x

2 = 0

This is a CONTRADICTION and hence we can see that 2 / 0 is not POSSIBLE ...

Which means that if we have 1÷x, this would be UNDEFINED if x = 0.

QUESTION -

If we have 1÷(x+1), for what value of x would render this undefinable?

For 1÷(x+1), it will be UNDEFINABLE when (x + 1) = 0 ... that means that solving it would make x = -1 ... hence, when x = -1, the function will be undefinable ... Consider the GRAPH of the function y = 1÷(x+1) (using grapher), what do you notice at x = -1?

What about 1÷(2x-3)? Hence, what would be general concept be that would make the expression 1÷(ax+b), undefinable? Does this change if the numerator was something else?

Hence, for functions of the form y = 1÷(ax+b), the DENOMINATOR can never be equal to ZERO ... and this will happen when x = -b ÷ a ... CHECK TO SEE THAT YOU KNOW WHY???

QUESTION -

In order to solve 2x - 5 = 3, we need to find the value of x, that SATISFIES (we discussed this in class today) the equation. Hence, we will have to leave x alone on the left-hand side. That means, we need to remove the constant (-5) and the coefficient (2) ...

How can this be done? I believe that Zhi Qi has a slight issue with this.

Well, when 2x - 5 = 3, we need to REMOVE the 5 on the LEFT ... hence,

2x - 5 + 5 = 3 + 5 ... NOTE that we have ADDED 5 on both sides ... leaving us with 2x = 8 ...

removing the coefficient 2 requires a DIVISION

2x ÷ 2 = 8 ÷ 2 ... leaving us with x = 4 ...

Hence, the EQUATION is SATISFIED when x = 4

CHECKING ... 2 (4 ) - 3 = 5 when is the same as the VALUE as provided in the question!

QUESTION -

When you're trying to SOLVE equations involving algebraic FRACTIONS, what is the general principal involved for questions of this form?

The general principal is to ensure that only 1 fraction is present on both sides of the equal sign!

Welcome Back for a New Semester

Dear students,

Welcome back to school after your June Holidays.

Let us start the new semester with this task.

Under the comment section, post up

1) 1 interesting thing you have done / see during the June Holidays.

2) 1 interesting knowledge that you have learn during the June Holidays.

3) The expectation that you are going to set for yourself in the learning of Mathematics.